In repeated computations of a large block matrix, I have noticed that a particular block is always symmetric.
It is the following matrix: $$ B\left[\left(I-\frac{h^2}{8}A^{-1}B\right)^{-1}+\frac12 I\right], $$ where $A$ and $B$ are both symmetric and positive-definite, $I$ is the identity matrix, and $h\in\mathbb{R}_{>0}$.
At first, I thought that this was trivial because $A$ and $B$ are both symmetric, and the other matrix involved is the identity.
However, taking a closer look, I fail to see why would this matrix reveal to be symmetric because the product $A^{-1}B$ is not symmetric. Therefore, neither the sum $I-\frac{h^2}{8}A^{-1}B$, nor its inverse, are symmetric. Then the product
$$ B\left(I-\frac{h^2}{8}A^{-1}B\right)^{-1} $$
has no reason to by symmetric. Or is it?
Is there a particular property that I am forgetting here? How can I prove that this block is always symmetric?
An initial simplification helps. Absorb $\frac{h^2}{8}$ into $B$ and forget about it.
Now note that since $B$ is symmetric, our given matrix is symmetric if and only if $X:=B(I-A^{-1}B)^{-1}$ is symmetric.
But $X$ is symmetric if and only if $X^{-1}$ is symmetric. As $(I-A^{-1}B)B^{-1}=B^{-1}(I-BA^{-1})$ we are done.